np.eye() for identity matrices in NumPy - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of np.eye() with default parameters

What is the output of this code snippet?

import numpy as np
print(np.eye(3))

NumPy

import numpy as np
print(np.eye(3))

A

[[0. 1. 0.]
 [1. 0. 1.]
 [0. 1. 0.]]

B

[[1. 0. 0.]
 [0. 1. 0.]
 [0. 0. 1.]]

C

[[1 0 0]
 [0 1 0]
 [0 0 1]]

D

[[0. 0. 0.]
 [0. 0. 0.]
 [0. 0. 0.]]

💡 Hint

np.eye(n) creates an n x n matrix with 1s on the main diagonal and 0s elsewhere.

np.eye(3) creates a 3x3 identity matrix with 1.0 on the diagonal and 0.0 elsewhere.

❓ Predict Output

intermediate

Effect of k parameter in np.eye()

What is the output of this code?

import numpy as np
print(np.eye(4, k=1))

NumPy

import numpy as np
print(np.eye(4, k=1))

A

[[0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]
 [0. 0. 0. 0.]]

B

[[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]

C

[[0. 0. 0. 0.]
 [1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]]

D

[[0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [1. 0. 0. 0.]]

💡 Hint

The k parameter shifts the diagonal by k positions above (positive) or below (negative) the main diagonal.

k=1 shifts the diagonal of 1s one step above the main diagonal, so the 1s appear just above the main diagonal.

❓ data_output

advanced

Shape of np.eye() with different rows and columns

What is the shape of the array created by this code?

import numpy as np
arr = np.eye(3, 5)
print(arr.shape)

NumPy

import numpy as np
arr = np.eye(3, 5)
print(arr.shape)

A (5, 3)

B (5, 5)

C (3, 5)

D (3, 3)

💡 Hint

The first argument is the number of rows, the second is the number of columns.

np.eye(3, 5) creates a matrix with 3 rows and 5 columns, so the shape is (3, 5).

❓ visualization

advanced

Visual pattern of np.eye() with negative k

Which option shows the correct printed output of this code?

import numpy as np
print(np.eye(4, k=-1))

NumPy

import numpy as np
print(np.eye(4, k=-1))

A

[[1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]]

B

[[0. 1. 0. 0.]
 [0. 0. 1. 0.]
 [0. 0. 0. 1.]
 [0. 0. 0. 0.]]

C

[[0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [0. 0. 0. 0.]
 [1. 0. 0. 0.]]

D

[[0. 0. 0. 0.]
 [1. 0. 0. 0.]
 [0. 1. 0. 0.]
 [0. 0. 1. 0.]]

💡 Hint

Negative k shifts the diagonal below the main diagonal.

k=-1 places 1s on the diagonal just below the main diagonal, so the 1s start at position (1,0).

🧠 Conceptual

expert

Understanding dtype effect in np.eye()

What will be the dtype and output of this code?

import numpy as np
arr = np.eye(2, dtype=int)
print(arr)
print(arr.dtype)

NumPy

import numpy as np
arr = np.eye(2, dtype=int)
print(arr)
print(arr.dtype)

A

[[1 0]
 [0 1]]
int64

B

[[1 0]
 [0 1]]
float64

C

[[1. 0.]
 [0. 1.]]
float64

D

[[1. 0.]
 [0. 1.]]
int32

💡 Hint

dtype=int forces the matrix elements to be integers instead of floats.

np.eye with dtype=int creates an integer identity matrix. The default int type on many systems is int64.

Practice

1. What does the function np.eye(4) create in NumPy?

easy

A. A 4x4 identity matrix with 1s on the main diagonal

B. A 4x4 matrix filled with zeros

C. A 4x4 matrix filled with ones

D. A 4x4 diagonal matrix with 1s on the first upper diagonal

Solution

1. Step 1: Understand np.eye() default behavior

   By default, np.eye(n) creates an n x n identity matrix with 1s on the main diagonal and 0s elsewhere.

2. Step 2: Apply np.eye(4)

   Calling np.eye(4) creates a 4x4 matrix with 1s on the main diagonal and 0s elsewhere.

3. Final Answer:

   A 4x4 identity matrix with 1s on the main diagonal -> Option A

4. Quick Check:

   np.eye(4) = 4x4 identity matrix [OK]

Hint: np.eye(n) makes nxn identity matrix by default [OK]

Common Mistakes:

• Confusing np.eye() with np.ones() or np.zeros()
• Thinking np.eye() fills all elements with 1
• Assuming 1s are on an off-diagonal by default

2. Which of the following is the correct syntax to create a 3x5 identity matrix with 1s on the main diagonal using np.eye()?

easy

A. np.eye(5, 3)

B. np.eye(3, 5)

C. np.eye(3)

D. np.eye(5)

Solution

1. Step 1: Understand np.eye() parameters

   The first parameter is the number of rows, the second is columns. So np.eye(3, 5) creates a 3x5 matrix.

2. Step 2: Check the options

   np.eye(3, 5) correctly uses np.eye(3, 5) to create a 3x5 matrix with 1s on the main diagonal.

3. Final Answer:

   np.eye(3, 5) -> Option B

4. Quick Check:

   np.eye(rows, cols) = correct syntax [OK]

Hint: First arg = rows, second = columns in np.eye() [OK]

Common Mistakes:

• Swapping rows and columns parameters
• Using only one parameter for non-square matrix
• Confusing np.eye() with np.identity()

3. What is the output of the following code?

import numpy as np
print(np.eye(3, k=1))

medium

A. [[0. 0. 1.]
[0. 1. 0.]
[1. 0. 0.]]

B. [[1. 0. 0.]
[0. 1. 0.]
[0. 0. 1.]]

C. [[0. 0. 0.]
[1. 0. 0.]
[0. 1. 0.]]

D. [[0. 1. 0.]
[0. 0. 1.]
[0. 0. 0.]]

Solution

1. Step 1: Understand the k parameter in np.eye()

   The parameter k=1 shifts the diagonal of 1s one above the main diagonal.

2. Step 2: Visualize the 3x3 matrix with 1s on the first upper diagonal

   The 1s appear at positions (0,1) and (1,2), zeros elsewhere.

3. Final Answer:

   [[0. 1. 0.] [0. 0. 1.] [0. 0. 0.]] -> Option D

4. Quick Check:

   k=1 shifts diagonal up by 1 [OK]

Hint: k shifts diagonal: 0=main, 1=above, -1=below [OK]

Common Mistakes:

• Ignoring the k parameter effect
• Assuming k=1 means main diagonal
• Confusing rows and columns in output

4. The code below throws an error. What is the mistake?

import numpy as np
matrix = np.eye(4, 3, k=4)
print(matrix)

medium

A. np.eye() does not accept three arguments

B. The matrix must be square for np.eye()

C. The value of k is too large for the matrix shape

D. The print statement is incorrect

Solution

1. Step 1: Check the meaning of k relative to matrix size

   The parameter k=4 means the diagonal of 1s is shifted 4 above the main diagonal.

2. Step 2: Compare k with matrix dimensions

   For a 4x3 matrix, the maximum upper diagonal shift is 2 (columns - 1). k=4 is invalid and causes an error.

3. Final Answer:

   The value of k is too large for the matrix shape -> Option C

4. Quick Check:

   k must be within valid diagonal range [OK]

Hint: k must fit within matrix diagonal limits [OK]

Common Mistakes:

• Assuming any k value works
• Thinking np.eye() only creates square matrices
• Blaming print instead of parameter error

5. You want to create a 5x5 matrix with 1s on the diagonal just below the main diagonal using np.eye(). Which code achieves this and what is the resulting matrix?

hard

A. np.eye(5, k=-1) resulting in 1s at positions (1,0), (2,1), (3,2), (4,3)

B. np.eye(5, k=1) resulting in 1s at positions (0,1), (1,2), (2,3), (3,4)

C. np.eye(5) resulting in 1s on the main diagonal

D. np.eye(5, k=-2) resulting in 1s two diagonals below the main

Solution

1. Step 1: Understand k parameter for diagonal shift

   Negative k shifts the diagonal of 1s below the main diagonal. k=-1 means one diagonal below.

2. Step 2: Apply np.eye(5, k=-1)

   This creates a 5x5 matrix with 1s at positions (1,0), (2,1), (3,2), (4,3) and zeros elsewhere.

3. Final Answer:

   np.eye(5, k=-1) with 1s just below main diagonal -> Option A

4. Quick Check:

   k=-1 shifts diagonal down by 1 [OK]

Hint: Use negative k for diagonals below main [OK]

Common Mistakes:

• Using positive k for below main diagonal
• Confusing positions of 1s in matrix
• Assuming np.eye() only creates main diagonal